$\log _2^x = 3 \Rightarrow x = {2^3} = 8$$\log _{\frac{1}{3}}^{x + 1} = \log _{\frac{1}{3}}^{8 + 1} = \log _{\frac{1}{3}}^9 = \log _{{3^{ - 1}}}^{{3^2}}$$ = \frac{2}{{ - 1}}\log _3^3 =  - 2 \times 1 =  - 2$